A136516 -id:A136516 - cp's OEIS frontend

A264871 Array read by antidiagonals: T(n,m) = (1+2^n)^m; n,m>=0.

1, 2, 1, 4, 3, 1, 8, 9, 5, 1, 16, 27, 25, 9, 1, 32, 81, 125, 81, 17, 1, 64, 243, 625, 729, 289, 33, 1, 128, 729, 3125, 6561, 4913, 1089, 65, 1, 256, 2187, 15625, 59049, 83521, 35937, 4225, 129, 1, 512, 6561, 78125, 531441, 1419857, 1185921, 274625, 16641, 257

Offset: 0

R. J. Mathar, Nov 27 2015

			       1,       2,       4,       8,      16,      32,
       1,       3,       9,      27,      81,     243,
       1,       5,      25,     125,     625,    3125,
       1,       9,      81,     729,    6561,   59049,
       1,      17,     289,    4913,   83521, 1419857,
       1,      33,    1089,   35937, 1185921,39135393,

Cf. A000079 (row 0), A000244 (row 1), A000351 (row 2), A001019 (row 3), A001026 (row 4), A009977 (row 5), A000051 (column 1), A028400 (column 2), A136516 (main diagonal), A165327 (upper subdiagonal).

Reverse /@ Table[(1 + 2^(n - m))^m, {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Nov 27 2015 *)

G.f. for row n: 1/(1-(1+2^n)*x). - R. J. Mathar, Dec 15 2015

A326011 a(n) = (n+1) * (2^n + 1)^n.

Original entry on oeis.org

1, 6, 75, 2916, 417605, 234812358, 527932234375, 4755738419928072, 171280331996409907209, 24606864966197875457438730, 14080929986159936046600341796875, 32073236633246852578917758577924120588, 290760173774986242601808360162358149769707533, 10492680499171055486742235424276666079725581443186702, 1507792223578968167717594884445653164343553232898773193359375

Offset: 0

Paul D. Hanna, Jun 05 2019

nonn

More generally, the following sums are equal:
(1) Sum_{n>=0} binomial(n+k-1, n) * y^n * (F + G^n)^n,
(2) Sum_{n>=0} binomial(n+k-1, n) * y^n * G^(n^2) / (1 - y*F*G^n)^(n+k),
for any fixed integer k; here, k = 2 and y = x, F = 1, G = 2.

			O.g.f.: A(x) = 1 + 6*x + 75*x^2 + 2916*x^3 + 417605*x^4 + 234812358*x^5 + 527932234375*x^6 + 4755738419928072*x^7 + ... + (n+1)*(2^n + 1)^n*x^n + ...
such that
A(x) = 1/(1 - x)^2 + 2*2*x/(1 - 2*x)^3 + 3*2^4*x^2/(1 - 2^2*x)^4 + 4*2^9*x^3/(1 - 2^3*x)^5 + 5*2^16*x^4/(1 - 2^4*x)^6 + 6*2^25*x^5/(1 - 2^5*x)^7 + 7*2^36*x^6/(1 - 2^6*x)^8 + ... + (n+1)*2^(n^2)*x^n/(1 - 2^n*x)^(n+2) + ...

Cf. A136516, A326012.

Table[(n+1)(2^n+1)^n,{n,0,20}] (* Harvey P. Dale, Mar 22 2020 *)

{a(n) = (n+1) * (2^n + 1)^n}
for(n=0,15, print1(a(n),", "))

/* O.g.f. */
{a(n) = my(A = sum(m=0,n, (m+1) * 2^(m^2) * x^m / (1 - 2^m*x +x*O(x^n))^(m+2) )); polcoeff(A,n)}
for(n=0,15, print1(a(n),", "))

/* E.g.f. */
{a(n) = my(A = sum(m=0,n, (m+1 + 2^m*x) * 2^(m^2) * exp(2^m*x +x*O(x^n)) * x^m/m! )); n!*polcoeff(A,n)}
for(n=0,15, print1(a(n),", "))

O.g.f.: Sum_{n>=0} (n+1) * (2^n + 1)^n * x^n.
O.g.f.: Sum_{n>=0} (n+1) * 2^(n^2) * x^n / (1 - 2^n*x)^(n+2).
E.g.f.: sum_{n>=0} (n+1 + 2^n*x) * 2^(n^2) * exp(2^n*x) * x^n/n!.

A326012 a(n) = (n+1)(n+2)/2 (2^n + 1)^n.

Original entry on oeis.org

1, 9, 150, 7290, 1252815, 821843253, 2111728937500, 21400822889676324, 856401659982049536045, 135337757314088315015913015, 84485579916959616279602050781250, 208476038116104541762965430756506783822, 2035321216424903698212658521136507048387952731, 78695103743782916150566765682074995597941860823900265, 12062337788631745341740759075565225314748425863190185546875000

Offset: 0

Paul D. Hanna, Jun 05 2019

nonn

More generally, the following sums are equal:
(1) Sum_{n>=0} binomial(n+k-1, n) * r^n * (p + q^n)^n,
(2) Sum_{n>=0} binomial(n+k-1, n) * r^n * q^(n^2) / (1 - p*q^n*r)^(n+k),
for any fixed integer k; here, k = 3 and p = 1, q = 2, r = x.

			O.g.f.: A(x) = 1 + 9*x + 150*x^2 + 7290*x^3 + 1252815*x^4 + 821843253*x^5 + 2111728937500*x^6 + 21400822889676324*x^7 + 856401659982049536045*x^8 + ... + (n+1)*(n+2)/2 * (2^n + 1)^n*x^n + ...
such that
A(x) = 1/(1 - x)^3 + 3*2*x/(1 - 2*x)^4 + 6*2^4*x^2/(1 - 2^2*x)^5 + 10*2^9*x^3/(1 - 2^3*x)^6 + 15*2^16*x^4/(1 - 2^4*x)^7 + 21*2^25*x^5/(1 - 2^5*x)^8 + 28*2^36*x^6/(1 - 2^6*x)^9 + ... + (n+1)*(n+2)/2 * 2^(n^2)*x^n/(1 - 2^n*x)^(n+3) + ...

Cf. A136516, A326011.

Table[((n+1)(n+2))/2 (2^n+1)^n,{n,0,20}] (* Harvey P. Dale, Jun 02 2025 *)

{a(n) = (n+1)*(n+2)/2 * (2^n + 1)^n}
for(n=0,15, print1(a(n),", "))

/* O.g.f. */
{a(n) = my(A = sum(m=0,n, (m+1)*(m+2)/2 * 2^(m^2) * x^m / (1 - 2^m*x +x*O(x^n))^(m+3) )); polcoeff(A,n)}
for(n=0,15, print1(a(n),", "))

/* E.g.f. */
{a(n) = my(A = sum(m=0,n, ((m+1 + 2^m*x)*(m+2 + 2^m*x) + 2^m*x)/2 * 2^(m^2) * exp(2^m*x +x*O(x^n)) * x^m/m! )); n!*polcoeff(A,n)}
for(n=0,15, print1(a(n),", "))

O.g.f.: Sum_{n>=0} (n+1)*(n+2)/2 * (2^n + 1)^n * x^n.
O.g.f.: Sum_{n>=0} (n+1)*(n+2)/2 * 2^(n^2) * x^n / (1 - 2^n*x)^(n+3).
E.g.f.: sum_{n>=0} ((n+1 + 2^n*x)*(n+2 + 2^n*x) + 2^n*x)/2 * 2^(n^2) * exp(2^n*x) * x^n/n!.

A337851 a(n) = (2^n + 2)^n.

Original entry on oeis.org

1, 4, 36, 1000, 104976, 45435424, 82653950016, 627485170000000, 19631688197463081216, 2504194578379511247798784, 1292628144912333835229805413376, 2687153475176994340820312500000000000, 22431765115399782718874449007331506546282496

Offset: 0

Paul D. Hanna, Sep 26 2020

nonn

In general, we have the o.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * x^n/(1 - b*q^n*x)^(n+1)  =  Sum_{n>=0} (m*q^n + b)^n * x^n ; here,  q=2, m=1, b=2.
In general, we have the e.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n / n!  =  Sum_{n>=0} (m*q^n + b)^n * x^n / n! ; here,  q=2, m=1, b=2.

			O.g.f.: A(x) = 1 + 4*x + 36*x^2 + 1000*x^3 + 104976*x^4 + 45435424*x^5 + 82653950016*x^6 + 627485170000000*x^7 + 19631688197463081216*x^8 + ...
where
A(x) = 1/(1 - 2*x) + 2*x/(1 - 2^2*x)^2 + 2^4*x^2/(1 - 2^3*x)^3 + 2^9*x^3/(1 - 2^4*x)^4 + 2^16*x^4/(1 - 2^5*x)^5 + 2^25*x^5/(1 - 2^6*x)^6 + ...

Cf. A165327, A055601, A251657, A337852, A136516.

{a(n,q,m,b) = (m*q^n + b)^n}
for(n=0,15, print1(a(n,q=2,m=1,b=2),", "))

/* E.g.f. formula: */
{a(n,q,m,b) = polcoeff( sum(k=0,n, m^k * q^(k^2) * x^k / (1 - b*q^k*x +x*O(x^n))^(k+1)), n)}
for(n=0,15, print1(a(n,q=2,m=1,b=2),", "))

/* E.g.f. formula: */
{a(n,q,m,b) = n! * polcoeff( sum(k=0,n, m^k * q^(k^2) * exp(b*q^k*x +x*O(x^n)) * x^k/k!), n)}
for(n=0,15, print1(a(n,q=2,m=1,b=2),", "))

O.g.f.: Sum_{n>=0} 2^(n^2) * x^n/(1 - 2^(n+1)*x)^(n+1)  =  Sum_{n>=0} (2^n + 2)^n * x^n.
E.g.f.: Sum_{n>=0} 2^(n^2) * exp(2^(n+1)*x) * x^n / n!  =  Sum_{n>=0} (2^n + 2)^n * x^n / n!.
a(n) = 2^n * A165327(n) for n >= 0.

A386644 E.g.f. A(x) satisfies A(x) = Sum_{n>=0} (A(x)^n + x)^n * x^n / n!.

Original entry on oeis.org

1, 1, 5, 34, 437, 7996, 191497, 5679178, 200959929, 8269303384, 388201586381, 20486491855534, 1201171090068325, 77504136748838164, 5460029344935045441, 417185040885539939506, 34377042102420367770353, 3040184386700809821194416, 287334696971272926921192469, 28915390444625255004763736278

Offset: 0

Paul D. Hanna, Aug 08 2025

nonn

It appears that for n > 6, a(n) (mod 6) equals [1, 4, 3, 4, 5, 4] repeating.
In general, the following sums are equal:
(C.1) Sum_{n>=0} (q^n + p)^n * r^n/n!,
(C.2) Sum_{n>=0} q^(n^2) * exp(p*q^n*r) * r^n/n!;
here, q = A(x) with p = x, r = x.

			E.g.f.: A(x) = 1 + x + 5*x^2/2! + 34*x^3/3! + 437*x^4/4! + 7996*x^5/5! + 191497*x^6/6! + 5679178*x^7/7! + 200959929*x^8/8! + 8269303384*x^9/9! + ...
where
A(x) = 1 + (A(x) + x)*x + (A(x)^2 + x)^2*x^2/2! + (A(x)^3 + x)^3*x^3/3! + (A(x)^4 + x)^4*x^2/4! + (A(x)^5 + x)^5*x^5/5! + ...
Also,
A(x) = exp(x^2) + A(x)*exp(x^2*A(x))*x + A(x)^4*exp(x^2*A(x)^2)*x^2/2! + A(x)^9*exp(x^2*A(x)^3)*x^3/3! + A(x)^16*exp(x^2*A(x)^4)*x^4/4! + ...

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A386645, A136516.

{a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); A[#A] = polcoeff( sum(m=0, #A, (Ser(A)^m + x)^m * x^m/m! ), #A-1) ); H=A; n!*A[n+1]}
for(n=0, 20, print1(a(n), ", "))

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = Sum_{n>=0} (A(x)^n + x)^n * x^n / n!.
(2) A(x) = Sum_{n>=0} A(x)^(n^2) * exp(x^2*A(x)^n) * x^n / n!.

A337852 a(n) = (2^(n+1) + 1)^n.

Original entry on oeis.org

1, 5, 81, 4913, 1185921, 1160290625, 4608273662721, 74051159531521793, 4796659837465472798721, 1248862969947666168212890625, 1304426412609681656861792686592001, 5459157240288132828933147334116110282753, 91477746675481294892349178081259839233191936001

Offset: 0

Paul D. Hanna, Sep 26 2020

nonn

In general, we have the o.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * x^n/(1 - b*q^n*x)^(n+1)  =  Sum_{n>=0} (m*q^n + b)^n * x^n ; here,  q=2, m=2, b=1.
In general, we have the e.g.f. identity:
Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n / n!  =  Sum_{n>=0} (m*q^n + b)^n * x^n / n! ; here,  q=2, m=2, b=1.

			O.g.f.: A(x) = 1 + 5*x + 81*x^2 + 4913*x^3 + 1185921*x^4 + 1160290625*x^5 + 4608273662721*x^6 + 74051159531521793*x^7 + 4796659837465472798721*x^8 + ...
where
A(x) = 1/(1 - x) + 2^2*x/(1 - 2*x)^2 + 2^6*x^2/(1 - 2^2*x)^3 + 2^12*x^3/(1 - 2^3*x)^4 + 2^20*x^4/(1 - 2^4*x)^5 + 2^30*x^5/(1 - 2^5*x)^6 + ...

Cf. A055601, A251657, A337851, A136516.

{a(n,q,m,b) = (m*q^n + b)^n}
for(n=0,15, print1(a(n,q=2,m=2,b=1),", "))

/* E.g.f. formula: */
{a(n,q,m,b) = polcoeff( sum(k=0,n, m^k * q^(k^2) * x^k / (1 - b*q^k*x +x*O(x^n))^(k+1)), n)}
for(n=0,15, print1(a(n,q=2,m=2,b=1),", "))

/* E.g.f. formula: */
{a(n,q,m,b) = n! * polcoeff( sum(k=0,n, m^k * q^(k^2) * exp(b*q^k*x +x*O(x^n)) * x^k/k!), n)}
for(n=0,15, print1(a(n,q=2,m=2,b=1),", "))

O.g.f.: Sum_{n>=0} 2^(n*(n+1)) * x^n/(1 - 2^n*x)^(n+1) = Sum_{n>=0} (2^(n+1) + 1)^n * x^n.

E.g.f.: Sum_{n>=0} 2^(n*(n+1)) * exp(2^n*x) * x^n / n! = Sum_{n>=0} (2^(n+1) + 1)^n * x^n / n!.

A386646 Expansion of e.g.f. Sum_{n>=0} (2^n + x)^n * x^n / n!.

Original entry on oeis.org

1, 2, 18, 536, 66316, 33636832, 68750980216, 562995064353920, 18446990378410477200, 2417856827427983647531520, 1267651025241922183470966470176, 2658456127743272591813667810372278272, 22300745369876426654206395965130496991176384, 748288839162767087393170357241926671150780067340288

Offset: 0

Paul D. Hanna, Aug 08 2025

nonn

In general, the following sums are equal:
(C.1) Sum_{n>=0} (q^n + p)^n * r^n/n!,
(C.2) Sum_{n>=0} q^(n^2) * exp(p*q^n*r) * r^n/n!;
here, q = A(x) with p = x, r = x.

			E.g.f.: A(x) = 1 + 2*x + 18*x^2/2! + 536*x^3/3! + 66316*x^4/4! + 33636832*x^5/5! + 68750980216*x^6/6! + 562995064353920*x^7/7! + 18446990378410477200*x^8/8! + ...
where
A(x) = 1 + (2 + x)*x + (2^2 + x)^2*x^2/2! + (2^3 + x)^3*x^3/n! + (2^4 + x)^4*x^4/4! + (2^5 + x)^5*x^5/5! + ...
Also,
A(x) = exp(x^2) + 2*exp(2*x^2)*x + 2^4*exp(2^2*x^2)*x^2/2! + 2^9*exp(2^3*x^2)*x^3/3! + 2^16*exp(2^4*x^2)*x^4/4! + ...

Paul D. Hanna, Table of n, a(n) for n = 0..70

Cf. A136516.

nmax = 15; CoefficientList[Series[Sum[(2^k + x)^k * x^k / k!, {k, 0, nmax}], {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Aug 09 2025 *)

{a(n) = my(A = sum(m=0, n, (2^m + x)^m * x^m/m! +x*O(x^n)) ); n!*polcoef(A,n)}
for(n=0, 15, print1(a(n), ", "))

E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies the following formulas.
(1) A(x) = Sum_{n>=0} (2^n + x)^n * x^n / n!.
(2) A(x) = Sum_{n>=0} 2^(n^2) * exp(2^n*x^2) * x^n / n!.
a(n) ~ 2^(n^2). - Vaclav Kotesovec, Aug 09 2025

A356274 a(n) is the number whose base-(n+1) expansion equals the binary expansion of n.

Original entry on oeis.org

1, 3, 5, 25, 37, 56, 73, 729, 1001, 1342, 1741, 2366, 2941, 3615, 4369, 83521, 104977, 130340, 160021, 194922, 234741, 280393, 332377, 406250, 474553, 551151, 636637, 732511, 837901, 954304, 1082401, 39135393, 45435425, 52521910, 60466213, 69345326, 79236613

Offset: 1

Thomas Scheuerle, Aug 02 2022

If the binary expansion of n is n = bit0*2^0 + bit1*2^1 + bit2*2^2 + ... then a(n) = bit0*(n+1)^0 + bit1*(n+1)^1 + bit2*(n+1)^2 + ... . In other words: Interpret the binary expansion of n as digits in base n+1.

			n=4 in binary is 100 and interpreting those digits as base n+1 = 5 is a(4) = 25.

Cf. A000523, A007814, A104257, A104258, A128889, A136516.

a[n_] := FromDigits[IntegerDigits[n, 2], n + 1]; Array[a, 40] (* Amiram Eldar, Aug 19 2022 *)

PARI
```
a(n) = fromdigits(digits(n, 2), n+1)
```

def a(n): return sum((n+1)**i*int(bi) for i, bi in enumerate(bin(n)[2:][::-1]))
print([a(n) for n in range(1, 39)]) # Michael S. Branicky, Aug 02 2022

a(2^n) = (2^n + 1)^n = A136516(n).
a(2^n - 1) = (2^(n^2) - 1)/(2^n - 1) = A128889(n).
a(2^n + 1) = 1 + (2^n + 2)^n.
a(n) = A104257(n+1, n).
a(n) = (1/n)*Sum_{j>=1} floor((n + 2^(j-1))/2^j) * ((n-1)*(n+1)^(j-1) + 1).
a(n) = (1/n)*Sum_{j=1..n} ((n-1)*(n+1)^A007814(j) + 1).
a(2*n) = A104258(2*n+1) - 1.
(2*m+1)^n divides a((2*m+1)^n-1) for positive m and n > 0.

cp's OEIS Frontend

A264871 Array read by antidiagonals: T(n,m) = (1+2^n)^m; n,m>=0.

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A326011 a(n) = (n+1) * (2^n + 1)^n.

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A326012 a(n) = (n+1)*(n+2)/2 * (2^n + 1)^n.

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A337851 a(n) = (2^n + 2)^n.

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A386644 E.g.f. A(x) satisfies A(x) = Sum_{n>=0} (A(x)^n + x)^n * x^n / n!.

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A337852 a(n) = (2^(n+1) + 1)^n.

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A386646 Expansion of e.g.f. Sum_{n>=0} (2^n + x)^n * x^n / n!.

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A326012 a(n) = (n+1)(n+2)/2 (2^n + 1)^n.